{"id":1466,"date":"2018-04-29T18:20:04","date_gmt":"2018-04-29T16:20:04","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1466"},"modified":"2022-11-19T15:16:29","modified_gmt":"2022-11-19T14:16:29","slug":"fs-domains-of-discs-and-formal-balls","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1466","title":{"rendered":"FS-domains of discs and formal balls"},"content":{"rendered":"

In his famous paper on the classification of continuous domains [1], Achim Jung showed that there were exactly two maximal categories of continuous dcpos: the L<\/strong>-domains, and the FS<\/strong>-domains. He invented the latter for the occasion, and they are the subject of exercises 9.6.25 through 9.6.32 in the book<\/a>.<\/p>\n

Every RB<\/strong>-domain is known to be an\u00a0FS<\/strong>-domain, and we know that all algebraic\u00a0FS<\/strong>-domains are RB<\/strong>-domains, in fact bifinite domains. The biggest conjecture about FS<\/strong>-domains is whether there is any\u00a0FS<\/strong>-domain at all that is not an\u00a0RB<\/strong>-domain. After 28 years, this is still an open problem.<\/p>\n

In [1], Achim finishes section 4 by “a concrete example of an FS<\/strong>-domain”, suggested to him by Jimmie Lawson. This is meant to be a likely candidate of a non-RB<\/strong> FS<\/strong>-domain… but we are lacking tools to show that something is not an RB<\/strong>-domain.<\/p>\n

FS-domains<\/h2>\n

In a dcpo X<\/em>, a Scott-continuous function f<\/em> : X<\/em> \u2192 X<\/em> is finitely separated from the identity<\/em> if and only if there is a finite subset M<\/em> of X<\/em> such that, for every x<\/em> \u2208 X<\/em>, there is an element m<\/em> \u2208 M<\/em> such that\u00a0f<\/em>(x<\/em>) \u2264\u00a0m<\/em> \u2264\u00a0x<\/em>.<\/p>\n

An\u00a0FS<\/strong>-domain is a dcpo\u00a0X<\/em> with a directed family (fi<\/sub><\/em>)i \u2208 I<\/sub><\/em> of Scott-continuous maps, finitely separated from the identity, and whose pointwise supremum is the identity map.<\/p>\n

For example, on [0, 1] with its usual ordering, for every \u03b5>0, the map\u00a0x<\/em>\u00a0\u21a6 max(x<\/em>\u2014\u03b5, 0) is finitely separated from the identity: take the integer multiples of\u00a0\u03b5 not exceeding 1 as separating set\u00a0M<\/em>, and maybe draw a picture to see what happens.<\/p>\n

When\u00a0\u03b5 varies, those functions\u00a0x<\/em>\u00a0\u21a6 max(x<\/em>\u2014\u03b5, 0) form a chain whose supremum is the identity map. \u00a0Hence [0, 1] is an\u00a0FS<\/strong>-domain. \u00a0It is much more of course. \u00a0For example, it is a continuous complete lattice, and all continuous complete lattices are bc-domains, all bc-domains are\u00a0RB<\/strong>-domains, and all\u00a0RB<\/strong>-domains are\u00a0FS<\/strong>-domains.<\/p>\n

The FS-domain of closed discs in the plane<\/h2>\n

Here is the example mentioned by Achim Jung in [1]. \u00a0Let\u00a0Disc<\/strong> be the poset of all closed disc in\u00a0R<\/strong>2<\/sup>, plus R<\/strong>2<\/sup>\u00a0itself, ordered by\u00a0reverse<\/em> inclusion. \u00a0Hence\u00a0R<\/strong>2<\/sup>\u00a0is the bottom element of\u00a0Disc<\/strong>.<\/p>\n

Achim mentions without proof that the filtered intersection of closed discs is a closed disc, and that is sufficiently obvious, geometrically, that I will avoid giving a formal proof of it. \u00a0Hence\u00a0Disc<\/strong> is a dcpo, and directed suprema are filtered intersections.<\/p>\n

Since every closed disc is compact and R<\/strong>2<\/sup>\u00a0is well-filtered (since sober, since Hausdorff),\u00a0Disc<\/strong> is a continuous dcpo, where a disc\u00a0D<\/em> is way-below another disc\u00a0D’<\/em> if and only if the interior of\u00a0D<\/em> contains\u00a0D’<\/em>. Hence the Scott topology on Disc<\/strong> coincides with the so-called upper Vietoris topology, whose basic open subsets are \u2610U<\/em> = {D<\/em> \u2208 Disc<\/strong> | D<\/em> \u2286 U<\/em>}, U<\/em> open in R<\/strong>2<\/sup>.<\/p>\n

For every\u00a0\u03b5>0, Achim\u00a0defines a map f\u03b5<\/sub><\/em> : Disc<\/strong> \u2192 Disc<\/strong> as follows: f\u03b5\u00a0<\/sub><\/em>maps every closed disc D<\/em>\u00a0inside the open ball B(0,<1\/\u03b5) with center 0 and radius 1\/\u03b5 to its\u00a0\u03b5-neighborhood, that is, to the disc with the same center and with radius equal to the radius of\u00a0D<\/em>, plus\u00a0\u03b5. \u00a0All other discs, and the whole plane itself, are mapped to the bottom element R<\/strong>2<\/sup>.<\/p>\n

This is really in the spirit of the function\u00a0x<\/em>\u00a0\u21a6 max(x<\/em>\u2014\u03b5, 0) we took to show that [0, 1] is an\u00a0FS<\/strong>-domain.<\/p>\n

Achim claims that f\u03b5<\/sub><\/em>\u00a0is Scott-continuous. It is easier to\u00a0prove that it is continuous with respect to the upper Vietoris topology, and to show that f\u03b5<\/sub><\/em>-1<\/sup> (\u2610U<\/em>) is open for every open subset U<\/em> of R<\/strong>2<\/sup>. That is clear if U<\/em> is the whole of R<\/strong>2<\/sup>, so let us assume it is not. Then f\u03b5<\/sub><\/em>-1<\/sup> (\u2610U<\/em>) is the set of closed discs D\u00a0<\/em>that are included in B(0,<1\/\u03b5) and in the “shrinking of U<\/em> by \u03b5”, U<\/em>-\u03b5<\/sup>, defined as the union of the open balls B(x<\/em>, <r<\/em>) such that the expanded ball B(x<\/em>, <r<\/em>+\u03b5) is included in U<\/em>. \u00a0(I hope this is correct. \u00a0I have not checked the details.)<\/p>\n

The fact that f\u03b5<\/sub><\/em> is finitely separated from the identity seems less elementary to me. Achim has a very short argument, which however appeals to a lot of untold facts: that the upper Vietoris topology is one half of the so-called Vietoris topology, that the latter, once restricted to a compact subset of R<\/strong>2<\/sup> (namely, the closed<\/em> ball of center 0 and radius 1\/\u03b5), is metrized by the so-called Hausdorff metric, etc.<\/p>\n

Lawson’s argument<\/h2>\n

Lawson [2] has a more elementary argument, and builds the separating set M<\/em> explicitly. He actually proves a more general theorem, but here is what his construction gives us in the present case. As I have just said, the closed ball with center 0 and radius 1\/\u03b5 is compact, hence it can be covered by finitely may open balls B(xi<\/sub><\/em>, <\u03b5\/3), 1\u2264i<\/em>\u2264k<\/em>.<\/p>\n

Here is a picture of the situation. \u00a0The small crosses are meant to be the set of points xi<\/sub><\/em>.<\/p>\n

\"\"<\/a><\/p>\n

Let M<\/em> be the set of closed discs with centers xi<\/sub><\/em> and radii \u03b5k<\/em>\/3, 2\u2264k<\/em>\u22643\/\u03b52<\/sup>+1, plus the bottom element R<\/strong>2<\/sup>. For every disc D<\/em>, if D<\/em> is not included in B(0, <1\/\u03b5), then, choosing m<\/em>=R<\/strong>2<\/sup>, we obtain f\u03b5<\/sub><\/em>(D<\/em>) \u2264\u00a0m<\/em> \u2264\u00a0D<\/em>: good (remember that \u2264 is reverse inclusion). Hence assume D<\/em>\u2286B(0, <1\/\u03b5).<\/p>\n

Otherwise, let x<\/em> be the center of D<\/em> and r<\/em> be its radius. \u00a0Here is a picture of\u00a0D<\/em>:<\/p>\n

\"\"<\/a><\/p>\n

We recall that, in that case, f\u03b5<\/sub><\/em>(D<\/em>) is the closed ball with center x<\/em> and radius\u00a0r<\/em>+\u03b5. \u00a0This is a smaller element of\u00a0Disc<\/strong>, but since the order is reverse inclusion, this is really a larger ball, geometrically:<\/p>\n

\"\"<\/a><\/p>\n

By construction, x<\/em> is in some open ball B(xi<\/sub><\/em>, <\u03b5\/3). \u00a0We have shown such an xi<\/sub><\/em>\u00a0in the above picture. \u00a0Let us look at the closed discs\u00a0centered at xi<\/sub><\/em>, and whose radii are multiples of \u03b5\/3, so that they fall in\u00a0M<\/em>. \u00a0Those are the discs shown with dashed lines below:<\/p>\n

\"\"<\/a><\/p>\n

As the picture shows, one those dashed discs contains\u00a0D<\/em> and is contained in f\u03b5<\/sub><\/em>(D<\/em>). \u00a0Let us show that formally.<\/p>\n

Since D<\/em>\u2286B(0, <1\/\u03b5), d(0,x<\/em>)+r<\/em> < 1\/\u03b5, where d(0,x<\/em>) is the distance from 0 to x<\/em>. That implies r<\/em> < 1\/\u03b5. Let k<\/em>\u22652 be such that \u03b5(k<\/em>-2)\/3\u2264r<\/em><\u03b5(k<\/em>-1)\/3. We check easily that k<\/em>\u22643\/\u03b52<\/sup>+1. We let\u00a0m<\/em>\u00a0\u2208\u00a0M<\/em> be the closed disc with center xi<\/sub><\/em>\u00a0and radius \u03b5k<\/em>\/3. \u00a0Then:<\/p>\n